Question

The number of positive integral solutions of $\tan^{-1} x + \cos^{-1} \left( \frac{y}{\sqrt{1+y^2}} \right) = \sin^{-1} \left( \frac{3}{\sqrt{10}} \right)$ are ______.

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Converting Diophantine fractions like $x = \frac{3y-1}{y+3}$ into mixed form $3 - \frac{10}{y+3}$ is a brilliant shortcut. It instantly restricts the infinite possibilities down to the simple factors of the constant numerator!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We must convert all inverse trigonometric terms into $\tan^{-1}$ to simplify the equation, then algebraically solve for the number of strictly positive integer pairs $(x, y)$ that satisfy it.

Step 2: Detailed Explanation:
Convert $\cos^{-1} \left( \frac{y}{\sqrt{1+y^2}} \right)$ to $\tan^{-1}$:
Let $\theta = \cos^{-1} \left( \frac{y}{\sqrt{1+y^2}} \right) \implies \cos \theta = \frac{y}{\sqrt{1+y^2}}$.
Using a right triangle where adjacent = $y$, hypotenuse = $\sqrt{1+y^2}$, the opposite side = $\sqrt{(\sqrt{1+y^2})^2 - y^2} = 1$.
Thus, $\tan \theta = \frac{\text{Opp}}{\text{Adj}} = \frac{1}{y} \implies \theta = \tan^{-1}\left(\frac{1}{y}\right)$.
Convert $\sin^{-1} \left( \frac{3}{\sqrt{10}} \right)$ to $\tan^{-1}$:
Let $\phi = \sin^{-1} \left( \frac{3}{\sqrt{10}} \right) \implies \sin \phi = \frac{3}{\sqrt{10}}$.
Using a right triangle where opposite = $3$, hypotenuse = $\sqrt{10}$, the adjacent side = $\sqrt{10 - 9} = 1$.
Thus, $\tan \phi = \frac{\text{Opp}}{\text{Adj}} = \frac{3}{1} = 3 \implies \phi = \tan^{-1}(3)$.
Substitute these back into the original equation:
$\tan^{-1} x + \tan^{-1}\left(\frac{1}{y}\right) = \tan^{-1}(3)$
Using the addition formula $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A+B}{1-AB}\right)$:
$\tan^{-1} \left( \frac{x + 1/y}{1 - x/y} \right) = \tan^{-1}(3)$
$\frac{xy + 1}{y - x} = 3$
Solve this algebraically for $x$:
$xy + 1 = 3(y - x)$
$xy + 1 = 3y - 3x$
Bring $x$ terms to one side:
$xy + 3x = 3y - 1$
$x(y + 3) = 3y - 1$
$x = \frac{3y - 1}{y + 3}$
To find integer solutions easily, perform polynomial division:
$x = \frac{3(y + 3) - 10}{y + 3} = 3 - \frac{10}{y + 3}$
For $x$ to be an integer, $(y + 3)$ must be a positive divisor of 10.
Divisors of 10 are: 1, 2, 5, 10.
Since $y$ must be a positive integer ($y \ge 1$), we have $(y + 3) \ge 4$.
Therefore, the only valid divisors for $(y + 3)$ are 5 and 10.
Case 1: $y + 3 = 5 \implies y = 2$.
Substitute $y=2$ to find $x$: $x = 3 - \frac{10}{5} = 3 - 2 = 1$.
Solution pair: $(1, 2)$.
Case 2: $y + 3 = 10 \implies y = 7$.
Substitute $y=7$ to find $x$: $x = 3 - \frac{10}{10} = 3 - 1 = 2$.
Solution pair: $(2, 7)$.
There are exactly 2 valid pairs of positive integral solutions.

Step 3: Final Answer:
The number of solutions is 2, matching option (b).